New year beginning

As the first four digits of $X^2$ is $2018$ , we can conclude that $2019×10^n>X^2 \geq 2018×10^n \\ \sqrt{2019×10^n}>X \geq \sqrt{2018×10^n}$

Case 1: $n=2k$ ( $n$ is even) $\sqrt{2019}×10^k>X \geq \sqrt{2018}×10^k \\ 44.933×10^k> X \geq 44.922×10^k \\ 4493.3×10^{k-2}> X \geq 4492.2×10^{k-2}$ For $k=0,1$ , there doesn't exist an integer which satisfy $X$ . So, the minimum value in Case 1 is when $k-2=0$ , and $X=4493$ .

Case 2: $n=2k+1$ ( $n$ is odd) $\sqrt{20190}×10^k>X \geq \sqrt{20180}×10^k \\ 142.09×10^k>X \geq 142.056×10^k \\ 14209×10^{k-2}> X \geq 14205.6×10^{k-2}$ In Case 2, the minimum value of $X$ is $14206$ .

Hence, the minimum value of $X$ is $\large 4493$ .

Firstly, note that the two closest integers to $\sqrt{2018}$ are $44$ and $45$ , and their squares are $1936$ and $2025$ .

Since there is no integer between these two that starts with $2018$ , then we can add an extra digit - testing out $441, 442, 443$ and so on. Since $1936$ is 'behind' $2018$ , but $2025$ is past $2018$ , then if there is a number that starts with $2018$ , it must lie between $440$ and $450$ . However, $449^2$ starts with $2016$ , and $450^2$ starts with $2025$ . Since we haven't found an integer that satisfies this, we choose the closest of the two - $449$ , and repeat the process again.

We now search the $4$ -digit integers between $4490$ and $4500$ . The first integer that satisfies this is $4493$ , which when squared is $20187049$ and clearly begins with $2018$ .